Q.Find the Median of the marks obtained by 5 students in an economics test: 15, 22, 9, 30, 18.

Showing the 12 most recent of 17 on this concept.

1. CA Foundation 2026Set jan-20261 markMCQ
   Q.If arithmetic mean of two numbers is 64 and harmonic mean is 16 then geometric mean is (A) 64 (B) 16 (C) 32 (D) 8

   ›Reveal solutionSolution

   Using the relation $GM^2 = AM\times HM$, $GM=\sqrt{64\times16}=\sqrt{1024}=32$.

   Step 1 — recall the relation

   For two positive numbers, $AM \times HM = GM^2$.

   Step 2 — substitute

   $GM^2 = 64 \times 16 = 1024$

   Step 3 — solve

   $GM = \sqrt{1024} = 32$

   Watch out

   The relation $GM^2=AM\times HM$ holds only for two numbers. Also resist just averaging 64 and 16 ($=40$) or picking one of the given values — the geometric mean is the square root of their product.

   Tip

   $GM$ is itself the geometric mean of $AM$ and $HM$: $GM=\sqrt{AM\times HM}=\sqrt{64\times16}$. Spotting $1024=2^{10}$ gives $\sqrt{1024}=2^5=32$ instantly.

   ✓Final answer

   (C) 32

2. CA Foundation 2025Set jan-20251 markMCQ
   Q.If the mode of the following data is 13, then the value of x in the data set is $13, 8, 6, 3, 8, 13, 2x + 3, 8, 13, 3, 5, 7$ (A) 6 (B) 5 (C) 7 (D) 8

   ›Reveal solutionSolution

   To make 13 the mode, set $2x+3=13\Rightarrow x=5$, which raises 13's count to four (beating 8's three).

   Step 1 — Count the fixed values

   Data: $13, 8, 6, 3, 8, 13, 2x+3, 8, 13, 3, 5, 7$.

   Ignoring the unknown term: $13$ appears $3$ times, $8$ appears $3$ times, $3$ appears $2$ times; all others once.

   Step 2 — Use the mode condition

   The mode is given as $13$. With $8$ also appearing three times, $13$ must occur MORE than $8$, so the unknown term must itself be $13$:

   $2x+3 = 13 \;\Rightarrow\; 2x = 10 \;\Rightarrow\; x = 5.$

   Now $13$ appears four times — the unique mode.

   Why the other options are wrong: $x=6\Rightarrow2x+3=15$, $x=7\Rightarrow17$, $x=8\Rightarrow19$ — none adds to the 13-count, leaving 8 and 13 tied (no single mode 13).

   Watch out

   Solve for the term VALUE $2x+3$ first, then back out $x$. Answering $x=13$ (forgetting the $2x+3$ expression) is the classic trap.

   Tip

   For a "make this the mode" question, force the unknown to equal the required modal value, then tally to confirm it wins outright.

   ✓Final answer

   (B) 5

3. CA Foundation 2025Set jan-20251 markMCQ
   Q.The best measure of central tendency is (A) Mean (B) Median (C) Mode (D) Range

   ›Reveal solutionSolution

   The arithmetic mean is the best measure of central tendency on the standard textbook criteria.

   Step 1 — Criteria for an ideal average

   An ideal measure of central tendency should be (i) rigidly defined, (ii) based on all observations, (iii) easy to understand and compute, (iv) capable of further algebraic treatment and (v) least affected by sampling fluctuations.

   Step 2 — Why the mean wins

   The arithmetic mean satisfies all these — every value enters the calculation and it feeds directly into higher statistics (combined mean, standard deviation, regression). It is therefore taken as the best/most widely used measure.

   Why the other options are wrong: the median and mode ignore much of the data and cannot be algebraically combined easily; the range (D) is not even a measure of central tendency — it measures spread.

   Watch out

   The mean's one weakness is sensitivity to extreme values; when outliers dominate, the median is preferred — but as a general "best" measure, the mean is the standard answer.

   Tip

   Eliminate the odd one out first: range is dispersion, so it is gone immediately.

   ✓Final answer

   (A) Mean

4. CA Foundation 2025Set may-20251 markMCQ
   Q.Which one of the following measures of central tendency is based on only fifty percent (50%) of the central values ? (A) Geometric Mean (B) Harmonic Mean (C) Median (D) Mode

   ›Reveal solutionSolution

   The median is the positional average determined by the central 50% of the ordered data — not by the extreme values.

   Step 1 — What each measure uses

   • Geometric mean and Harmonic mean are mathematical averages computed from all the observations.
   • Mode is the value with the highest frequency — fixed by the most concentrated value.
   • Median is a positional average: arrange the data and take the middle value, so it is governed by where the central items fall, splitting the series into two 50% halves.

   Step 2 — Match the description

   “Based on only fifty percent (50%) of the central values” points to the average that is located by the central position and is unaffected by the magnitudes at the extremes — the median.

   Watch out

   The geometric and harmonic means use every observation, so they cannot be “based on 50% of central values.” Do not pick them on this description.

   Tip

   Median = positional, robust to extreme values. That robustness is exactly the “central values” property being described.

   ✓Final answer

   (C) Median

5. CA Foundation 2025Set may-20251 markMCQ
   Q.Find out the mode from the following data : 100, 110, 125, 225, 325, 125, 90, 80, 455, 375, 125 (A) 325 (B) 110 (C) 455 (D) 125

   ›Reveal solutionSolution

   The most frequently occurring value is 125 (it appears 3 times), so the mode is 125.

   Step 1 — Tally the frequencies

   Data: 100, 110, 125, 225, 325, 125, 90, 80, 455, 375, 125.

   Value	Times it appears
   125	3
   every other value (100, 110, 225, 325, 90, 80, 455, 375)	1 each

   Step 2 — Pick the mode

   The mode is the value with the highest frequency. Here 125 occurs three times, more than any other value.

   Watch out

   The mode is about frequency, not size. Do not pick the largest value (455) or a middle value — count how often each value repeats.

   Tip

   For raw ungrouped data, just scan for the value that repeats most. One clear repeat-leader = the mode.

   ✓Final answer

   (D) 125

6. CA Foundation 2024Set sep-20241 markMCQ
   Q.The Median of the following frequency distribution is :

   $x$	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
   $f(x)$	8	30	40	12	10
   (A) 33
   (B) 22.5
   (C) 23
   (D) 24

   ›Reveal solutionSolution

   Median class is 20–30 (where cumulative frequency crosses N/2 = 50); the formula gives 23.

   Step 1 — Cumulative frequencies ($N = 8+30+40+12+10 = 100$)

   Class	$f$	Cumulative $f$
   0–10	8	8
   10–20	30	38
   20–30	40	78
   30–40	12	90
   40–50	10	100

   Step 2 — Locate the median class

   $N/2 = 50$. The cumulative frequency first exceeds 50 in the 20–30 class (cf jumps 38 → 78), so median class = 20–30.

   Step 3 — Apply the median formula

   $M = L + \frac{\tfrac{N}{2} - cf}{f}\times h$

   with $L=20,\; cf=38,\; f=40,\; h=10$:

   $M = 20 + \frac{50 - 38}{40}\times 10 = 20 + \frac{12}{40}\times 10 = 20 + 3 = 23$

   Reject the others

   • (B) 22.5, (D) 24 → arise from wrong $cf$ or $L$.
   • (A) 33 → uses the wrong class.

   Watch out

   Use the cumulative frequency of the class before the median class ($cf=38$), not the median class's own frequency.

   ✓Final answer

   (C) 23

7. CA Foundation 2023Set jun-20231 markMCQ
   Q.A Professor has given assignment to students in a Statistics class. A student Jagan computes the arithmetic mean and standard deviation for a set of 100 observations as 50 and 5 respectively. Later on, Sonali points out to Jagan that he has made a mistake in taking one observation as 100 instead of 50. What would be the correct mean if the wrong observation is corrected? (A) 50.5 (B) 49.9 (C) 49.5 (D) 50.1

   ›Reveal solutionSolution

   Corrected sum 4950 over 100 observations gives a mean of 49.5.

   Step 1 — Recover the original total

   $\sum x = n\bar{x} = 100 \times 50 = 5000$

   Step 2 — Correct the wrong observation

   Replace 100 by 50:

   $\text{corrected sum} = 5000 - 100 + 50 = 4950$

   Step 3 — New mean

   $\bar{x}_{\text{corrected}} = \frac{4950}{100} = 49.5$

   Watch out

   The standard deviation of 5 is irrelevant to the mean correction — including it is a common trap.

   Tip

   Shortcut: the mean changes by (correction ÷ n) = $(50-100)/100 = -0.5$, so $50 - 0.5 = 49.5$.

   ✓Final answer

   (C) 49.5

   Note

   This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

8. CA Foundation 2023Set jun-20231 markMCQ
   Q.Find the mean of the following data

   Class interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
   Frequency	9	13	6	4	6	2	3
   (A) 23.7
   (B) 35.7
   (C) 39.7
   (D) 43.7

   ›Reveal solutionSolution

   Σfx = 1535, Σf = 43, so mean = 1535/43 ≈ 35.7.

   Step 1 — Class midpoints

   $15,\;25,\;35,\;45,\;55,\;65,\;75$

   Step 2 — Compute Σf and Σfx

   $\sum f = 9+13+6+4+6+2+3 = 43$

   $\sum fx = 135+325+210+180+330+130+225 = 1535$

   Step 3 — Mean

   $\bar{x} = \frac{\sum fx}{\sum f} = \frac{1535}{43} \approx 35.7$

   Watch out

   Divide by the total frequency (43), never by the number of class intervals (7) — that mistake inflates the mean badly.

   Tip

   A step-deviation method (assumed mean 45, common width 10) speeds up the arithmetic for grouped means.

   ✓Final answer

   (B) 35.7

   Note

   This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

9. CA Foundation 2023Set jun-20231 markMCQ
   Q.The Median of the following set of observations: 24, 18, 36, 42, 30, 28, 21, 29, 25, 33 is (A) 26.5 (B) 27.5 (C) 28.5 (D) 29.5

   ›Reveal solutionSolution

   Sorted, the middle pair (5th and 6th of 10) is 28 and 29 → median 28.5.

   Step 1 — Sort the data

   $18,\;21,\;24,\;25,\;28,\;29,\;30,\;33,\;36,\;42$

   Step 2 — Locate the median position

   With $n = 10$ (even), the median is the mean of the $\tfrac{n}{2}=5$th and $\tfrac{n}{2}+1=6$th values: 28 and 29.

   Step 3 — Compute

   $\text{Median} = \frac{28 + 29}{2} = 28.5$

   Watch out

   Never read the median off the unsorted list — always arrange the values in ascending order first.

   Tip

   For even n, the median is the average of the two central terms; for odd n it is the single middle term.

   ✓Final answer

   (C) 28.5

   Note

   This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

10. CA Foundation 2023Set jun-20231 markMCQ
    Q.Find the mode of the following data:

    X	25-30	30-35	35-40	40-45	45-50	50-55
    $f(x)$	20	53	42	42	41	43
    (A) 31.75
    (B) 30.75
    (C) 33.75
    (D) 35.75

    ›Reveal solutionSolution

    Modal class 30-35; mode = 30 + (33/44)×5 = 33.75.

    Step 1 — Identify the modal class

    Highest frequency is 53, in class 30-35. So $L = 30$, $f_1 = 53$, previous $f_0 = 20$, next $f_2 = 42$, width $h = 5$.

    Step 2 — Apply the mode formula

    $\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\times h$

    $= 30 + \frac{53-20}{2(53)-20-42}\times 5 = 30 + \frac{33}{44}\times 5$

    Step 3 — Compute

    $= 30 + 0.75\times 5 = 30 + 3.75 = 33.75$

    Watch out

    $f_0$ and $f_2$ are the frequencies of the classes immediately before and after the modal class — mixing them up (e.g., using 42 as $f_0$) changes the answer.

    Tip

    Denominator $2f_1 - f_0 - f_2 = (f_1-f_0)+(f_1-f_2)$; computing it as the sum of the two jumps reduces slip-ups.

    ✓Final answer

    (C) 33.75

    Note

    This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

11. CA Foundation 2023Set jun-20231 markMCQ
    Q.For a given data set: 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4; what is the median? (A) 8 (B) 6 (C) 4 (D) 9

    ›Reveal solutionSolution

    Sorted, the 8th of 15 values is the median = 6.

    Step 1 — Sort the data

    2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19.

    Step 2 — Locate the middle position

    $\text{Median position} = \frac{n+1}{2} = \frac{15+1}{2} = 8$

    The 8th value is 6.

    Watch out

    4 occurs most often (the mode) and tempts as an answer — but the median is the positional middle, not the most frequent value.

    Tip

    For odd n, the median is simply the ((n+1)/2)-th sorted term — no averaging needed.

    ✓Final answer

    (B) 6

    Note

    This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

12. CA Foundation 2022Set dec-20221 markMCQ
    Q.The median of the observations 42, 72, 35, 92, 67, 85, 72, 81, 51, 56 is (A) 69.5 (B) 72 (C) 64 (D) 61.5

    ›Reveal solutionSolution

    Sorted, the average of the 5th and 6th values is $(67+72)/2 = 69.5$.

    Step 1 — Arrange in ascending order

    $35,\;42,\;51,\;56,\;67,\;72,\;72,\;81,\;85,\;92$

    Step 2 — Locate the middle for even $n$

    With $n=10$, the median is the mean of the $\frac{n}{2}=5$th and $\frac{n}{2}+1=6$th observations.

    $\text{5th}=67,\qquad \text{6th}=72$

    Step 3 — Average them

    $\text{Median} = \frac{67+72}{2} = \frac{139}{2} = 69.5$

    Watch out

    You must sort first. Reading the raw list's middle entries (67, 85) or picking 72 because it repeats gives a wrong median.

    Tip

    Even $n$: average the two central sorted values; odd $n$: take the single central value.

    ✓Final answer

    (A) 69.5

    Note

    This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.